Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char alpha[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                  'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                  's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
string s;
int t;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  int tab[n + 1][m + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> tab[i][j];
    }
  }
  vector<pair<int, int>> r, c;
  int ans = 0;
  if (n < m) {
    for (int i = 1; i <= n; i++) {
      bool hasZero = false;
      int mx = INT_MAX;
      for (int j = 1; j <= m; j++) {
        if (tab[i][j] == 0) hasZero = true;
        mx = min(mx, tab[i][j]);
      }
      if (hasZero) continue;
      for (int j = 1; j <= m; j++) {
        tab[i][j] -= mx;
      }
      r.push_back({i, mx});
      ans += mx;
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 2; j <= n; j++) {
        if (tab[j][i] != tab[j - 1][i]) {
          cout << -1 << '\n';
          return 0;
        }
      }
      if (tab[1][i] != 0) {
        c.push_back({i, tab[1][i]});
        ans += tab[1][i];
      }
    }
  } else {
    for (int i = 1; i <= m; i++) {
      bool hasZero = false;
      int mx = INT_MAX;
      for (int j = 1; j <= n; j++) {
        if (tab[j][i] == 0) hasZero = true;
        mx = min(mx, tab[j][i]);
      }
      if (hasZero) continue;
      for (int j = 1; j <= n; j++) {
        tab[j][i] -= mx;
      }
      c.push_back({i, mx});
      ans += mx;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 2; j <= m; j++) {
        if (tab[i][j] != tab[i][j - 1]) {
          cout << -1 << '\n';
          return 0;
        }
      }
      if (tab[i][1] != 0) {
        r.push_back({i, tab[i][1]});
        ans += tab[i][1];
      }
    }
  }
  cout << ans << '\n';
  for (auto i : r) {
    for (int j = 0; j < i.second; j++) {
      cout << "row " << i.first << '\n';
    }
  }
  for (auto i : c) {
    for (int j = 0; j < i.second; j++) {
      cout << "col " << i.first << '\n';
    }
  }
}